Let $G$ is an abelian infinte group such that for all nontrivial subgroups $H$ $$\forall H\leq G, \left|\frac{G}{H}\right|<\infty$$ Prove that $G\cong\mathbb Z$.

What I have done:

Clearly, it is enough to show that $G$ is a cyclic group. Moreover, I see as $G/H$ is a finite group for all subgroups $H$, then $G$ cannot have any elements with the finite order. This means to me that if $H \leq G$ and $H$ is cyclic then $H$ cannot be finite. Help your friend for the rest. Thanks. :)

where $n_i \in \mathbb{Z}$. There is an isomorphic copy of each factor of the direct product in $G$. Quotienting $G$ by $\mathbb{Z}^s$ for various $1 \leq s \leq r$ and the $\mathbb{Z} / n_i \mathbb{Z}$, you see that $r = 1$ ($r \geq 1$ for the group to be infinite and $r \leq 1$ for the finite quotient property) and $k = 0$ (for the finite quotient property).

Your group is torsion-free (otherwise it has a finite subgroup of finite index, contradicting $G$ being infinite). Now let $g\in G$, and consider $H=\langle g\rangle$, which is of some finite index $n=[G:H]$. The map $G\rightarrow H$, $g\mapsto g^n$, then maps $G$ into an infinite cyclic group. The kernel, if non-trivial, would have finite index, impossible because $H$ is torsion-free. Thus the kernel is trivial and $G$ embeds in a cyclic group, so is itself cyclic.

This is indeed the better answer. William's answer is incomplete because your group is not necessarily finitely-generated.
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Chris GerigAug 26 '12 at 19:55

@ChrisGerig: thanks for the kind words, but the group will be finitely generated, because it is virtually cyclic (any subgroup generated by one element has finite index). I also prefer my answer :), because it avoids the classification.
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user641Aug 26 '12 at 20:52

1

This is beautiful in its simplicity.
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David WheelerAug 27 '12 at 22:12